Problem: $f(x)=\begin{cases} \text{sin}(x\cdot\pi)&\text{for }-8<x<0 \\\\ \dfrac x5&\text{for }0\leq x\leq10 \end{cases}$ Find $\lim_{x\to -5}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The limit doesn't exist.
Let's find the limit as $x$ approaches $-5$. We will use the fact that $f(x)=\text{sin}(x\cdot\pi)$ for $x$ -values between $-8$ and $0$. $\begin{aligned} &\phantom{=}\lim_{x\to -5}f(x) \\\\ &=\lim_{x\to -5}\text{sin}(x\cdot\pi) \\\\ &=\text{sin}(-5\pi)&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ In conclusion, we found that $\lim_{x\to -5}f(x)=0$.